Question: Find $\dfrac{d}{dx}\left[3x^2\sqrt{5x+3}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{6x}{2\sqrt{5x+3}}$ (Choice B) B $\dfrac{30x}{2\sqrt{5x+3}}$ (Choice C) C $\dfrac{75x^2+36x}{2\sqrt{5x+3}}$ (Choice D) D $\dfrac{3x^2+5x+3}{2\sqrt{5x+3}}$
Explanation: $3x^2\sqrt{5x+3}$ is a product of a function and a composite function. Let... $u(x)=3x^2$ $v(x)=\sqrt{x}$ $w(x)=5x+3$... then $3x^2\sqrt{5x+3}=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $\dfrac{d}{dx}\left[3x^2\sqrt{5x+3}\right]$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=6x$ $v'(x)=\dfrac{1}{2\sqrt{x}}$ $w'(x)=5$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=6x\cdot{\sqrt{5x+3}}+3x^2\cdot{\dfrac{1}{2\sqrt{5x+3}}}\cdot5 \\\\ &=6x\sqrt{5x+3}+\dfrac{15x^2}{2\sqrt{5x+3}} \\\\ &=\dfrac{12x(5x+3)}{2\sqrt{5x+3}}+\dfrac{15x^2}{2\sqrt{5x+3}} \\\\ &=\dfrac{75x^2+36x}{2\sqrt{5x+3}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[3x^2\sqrt{5x+3}\right]=\dfrac{75x^2+36x}{2\sqrt{5x+3}}$.